RIIT Major Test -3 Mains Solutions (18!01!2015) (1)

Batch: Juniors-16RIIT MAJOR TEST- 3

Main Level

ANSWER SHEET

PART-A: MATHEMATICS

1. (B) 2. (C) 3. (A) 4. (C) 5. (B)6. (B) 7. (C) 8. (C) 9. (C) 10. (A)11. (B) 12. (C) 13. (B) 14. (C) 15. (C)16. (D) 17. (B) 18. (A) 19. (C) 20. (C)21. (C) 22. (C) 23. (C) 24. (C) 25. (B)26. (B) 27. (C) 28. (A) 29. (B) 30. (B)

PART-B: PHYSICS

31. (D) 32. (D) 33. (A) 34. (B) 35. (A)36. (B) 37. (B) 38. (A) 39. (A) 40. (B)41. (B) 42. (C) 43. (D) 44. (A) 45. (C)46. (C) 47. (B) 48. (D) 49. (A) 50. (B)51. (C) 52. (A) 53. (A) 54. (B) 55. (B)56. (C) 57. (D) 58. (C) 59. (A) 60. (B)

PART-C: CHEMISTRY

61. (A) 62. (C) 63. (B) 64. (D) 65. (C)66. (B) 67. (A) 68. (B) 69. (A) 70. (C)71. (D) 72. (C) 73. (D) 74. (B) 75. (B)76. (B) 77. (B) 78. (C) 79. (B) 80. (C)81. (C) 82. (B) 83. (B) 84. (A) 85. (B)86. (B) 87. (B) 88. (C) 89. (B) 90. (A)

Date: 18-01-15

RAO IIT ACADEMY / JEE-Target -2016 Batch / Major Test -3 / Main Level / Solutions

2222SANTACRUZ | ANDHERI | GOREGAON | KANDIVALI (E) | KANDIWALI (W) | BORIVALI | BHAYANDER | VASAI | POWAI | DADAR |

SION | THANE | LOKPURAM (THANE) | DOMBIVLI | KALYAN | PANVEL | KAMOTHE | NERUL | SANPADA | KHARGHAR |

PART-A: MATHEMATICS= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =1. (B)Sol: Let f (x) = ax2 + 2bx – 5c

9ca b c4

4a + 4b + 4c > 9c 4a + 4b – 5c > 0 f (2) > 0 Equation f (x) = 0 has non-real roots then f (x) is either always +ve if a > 0 or always –ve if a < 0. f (2) > 0 a > 0Also f (0) > 0 [ f (x) is always +ve] –5c > 0 c < 0

Topic: Quadratic Equations_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_AGm sir.

2. (C)Sol: sec2 (a + 2)x + a2 – 1 = 0

tan2 (a + 2) x + a2 = 0 tan2 (a + 2)x = 0 and a = 0

tan2 (2x) = 0 x 0, ,2 2

Ordered pairs are

0,0 , 0, , 0,2 2

Topic: Trigonometric Equations_L-2_IIT-JEE_JR_Target-2016_MT-3_Main Level_AGm sir.

3. (A)Sol: Let variable circle be

x2 + y2 + 2gx + 2fy + c = 0So its center is (–g, –f) It touches x-axis i.e. y = 0

2 21 f1 g f c

f2 = g2 + f2 – c = 0 g2 – c = 0Also circle touches x2 + (y – 1)2 = 1 externally Distance b/w centers = sum of radius

22 2g 0 f 1 1 f

2 2 2g f 2f 1 1 f 2 f

2g 2 f 2f




Locus of center (–g, –f)

2g 2 f 2 f

2x 2 y 2y

When y 0 x2 = 4yWhen y < 0 x2 = – 2y + 2y = 0 x = 0

Locus is 2x 4y 0, y y 0

Topic: Circles_L-2_IIT-JEE_JR_Target-2016_MT-3_Main Level_AGm sir.

4. (C)Sol: Since a, b, c are in AP

2b = a + c 2 sin B = sin A + sin C

B B B A c4sin cos 2cos cos2 2 2 2

B B B A C4sin cos 2cos cos2 2 2 2

B 1 A C 1sin cos2 2 2 2

B 302 B 60

Topic: Solution of Triangle_L-2_IIT-JEE_JR_Target-2016_MT-3_Main Level_AGm sir.

5. (B)Sol: Equations of any line passing through (1, 1) can be given by y – 1 = m(x –1)

x – intercept 11m

y – intercept 1 m

Where and are roots of required eqn

Area of with coordinate axes 1 A2

2A

11 1 m 2Am

11 m 1 2Am

1m 2 2Am




Sum of roots 1 11 1 m 2 mm m

= 2 – (2 – 2A) = 2A equation will bex2 – 2Ax + 2A = 0

Topic: Quadratic Equations/ Straight Lines_L-2_IIT-JEE_JR_Target-2016_MT-3_Main Level_AGm sir.

6. (B)Sol: lines are concurrent

a b cb c a 0c a b

– (a3 + b3 + c3 – 3abc) = 0

(a + b + c) [a2 + b2 + c2 – ab –bc –ca] = 0

a b c 0 as a, b, c are sides of a2 + b2 + c2 – ab – bc – ca = 0

2 2 21 a b b c c a 02

a = b = c

equilateral Topic: Straight Lines_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_AGm sir.

7. (C)Sol: (ABA)T = { A(BA)}T = (BA)T AT

= ATBTAT

A and B are symmetric AAT = A BT = B (ABA)T = ABA ABA is also symmetric

Topic: Matrices_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_AGm sir.

8. (C)

Sol:n 11LHS

1

(sum of GP)

RHS Multiply 1 to numerator and denominator

2 4 8 161 1 1 1 1 11

2 2 4 8 161 1 1 1 1

1




321

1

LHS = RHS

n 1 321 1

1 1

n + 1 = 32 n = 31

Topic: Sequence & Series_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_AGm sir.

9. (C)Sol: is root of equation 4x2 + 2x – 1 = 0

24 2 1 0

3 24 2 0 3 24 2

3 32 f 2 4 3 1 2 4 2 1

2 22 2 2 1 4 2 2

24 2 1 1 = 0 + 1 =1

Topic: Quadratic Equations_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_AGm sir.

10. (A)Sol: Sloving the two lines we get

5x3 4m

x is an integer 3 + 4m = 1, –1, 5, –5

2 4 2 8m , , ,4 4 4 4

m has 2 integral valuesTopic: Straight Lines_L-2_IIT-JEE_JR_Target-2016_MT-3_Main Level_AGm sir.

11. (B)

Sol: 1/ 72 3 2

2 3 2

a a b b b c ca b c2 2 3 3 3 2 2

7 2 3 2

1/ 72 3 2

4 3

3 a b c7 2 3

10 4

2 3 27

3 2a b c7




Maximum value 10 4

7

3 27


12. (C)Sol: Rearranging the given equation

(a2 – 2a + 1) + (b2 – 2b + 1) + (c2 – 2c + 1) + (a2 – 2ab + b2) + (b2 – 2bc + c2) + (c2 – 2ca + a2) = 0 (a – 1)2 + (b – 1)2 + (c – 1)2 + (a – b)2 + (b – c)2 + (c – a)2 = 0 a = b = c = 1


13. (B)

Sol:

nn n 1 n 2

S3

n 1

n 1 n n 1S

3

n n n 1n n 1

t S S n 2 n 13

= n(n + 1)

n n

nr 1 r 1 r 1

1 1 1 1Str n n 1 n n 1

11

n 1

S 1 Topic: Sequence & Series_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_AGm sir.

14. (C)

Sol: ABC

x + y = 1

y = m x1

y = m x2

y = m x3

1

1 1

1 mA ,1 m 1 m

2

2 2

1 mB ,1 m 1 m

3

3 3

1 mC ,1 m 1 m

B is mid-point of A and C




2 1 2

2 1 11 m 1 m 1 m

1 + m1, 1 + m2, 1 + m3 are in HP

Note :

2 1 3

2 1 3

2m m m1 m 1 m 1 m

2 1 3

1 1 12 1 1 11 m 1 m 1 m

2 1 3

2 1 11 m 1 m 1 m

1 + m1, 1 + m2, 1 + m3 are in HP

Topic: Straight Lines_L-2_IIT-JEE_JR_Target-2016_MT-3_Main Level_AGm sir.

15. (C)

Sol: Circles are orthogonal 2

1 2 1 2 1 2c c 2 g g f f

line makes 2

with y-axis

It will is parallel to x-axis y – 2 = 0 y = 2

Topic: Circles_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_AGm sir.

16. (D)Sol: y = cos x cos (x + 2) – cos2 (x + 1)

y = cos (x + 1 – 1) cos (x + 1 + 1) – cos2 (x + 1)y = cos2 (x + 1) – sin21 – cos2 (x + 1)y = –sin21 line is parallel to x-axis passing through (k, –sin21) where k can be any value.

Topic: Straight Lines_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_AGm sir.

17. (B)

Sol: Any point on x + 2y = 1 can be taken as 1 kk,2

Chord of contact of 1 kk,2

on circle

x2 + y2 = 4 can be taken as

1 kxk y 42




y yx k 4 02 2

represents family of lines passing through point of intersection of lines yx 02

and y 4 02

i.e (4, 8)Topic: Circles_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_AGm sir.

18. (A)Sol: x2 – y2 + 3x + y + 2 = 0

(x – y + 2) (x + y + 1) = 0 The two diameters are x + y + 1 = 0 and x – y + 2 = 0

Center = Point of intersection of diameters 3 1,2 2

Circle passes through 1 , 12

Radius 2 23 1 1 51

2 2 2 2

Required circle is 4x2 + 4y2 + 12x – 4y – 15 = 0Topic: Circles_L-2_IIT-JEE_JR_Target-2016_MT-3_Main Level_AGm sir.

19. (C)Sol: Let P (x1, y1) Q (x2, y2) R (x3, y3)

Equation of chord of contacts will be xx1 + yy1 – c2 = 0 xx2 + yy2 – c2 = 0 xx3 + yy3 – c2 = 0 Lines are concurrent then

21 1 1 1

22 2 2 2

23 3 3 3

x y c x y 1x y c 0 or x y 1 0x y c x y 1

P, Q, R are collinearTopic: Circle_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_AGm sir.




20. (C)Sol: In a GP (tp)

2 = tp + q × tp – q

b2 = acDiseriminant of f(x) = 0 (2b)2 = 4 ac 4b2 – 4 ac = 0 af x 0 as f x 0 if a 0

or f x 0 if a 0


21. (C)

Sol:2

2

tan 1 3tanatan 3 3 tan

2 1 3atan 0

3 1

a > 3 or 1a3

Slope of 2

2

aOP aa

Slope of OP > 3 or 13

P lies in obtuse angle b/w the lies y = 3x and 3y = xTopic: Straight Lines_L-2_IIT-JEE_JR_Target-2016_MT-3_Main Level_AGm sir.

22. (C)Sol: Common chord of the circles will pass through center of 1st circle common chord is

10x +4y – a – b = 0Center of 1st circle = (– 4, – 4) – 40 – 16 – a – b = 0 a + b = – 56

Topic: Circle_L-2_IIT-JEE_JR_Target-2016_MT-3_Main Level_AGm sir.

23. (C)

Sol: We have 2 2

1 sin 1 sin is base of logarithm 0 sin 1 sinlog cos 2 2

2cos 2 sin

2 21 2sin sin

1sin3

0 sin 1

The given solution has a unique solutionTopic: Trigonometric Equations_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_AGm sir.


101010 10SANTACRUZ | ANDHERI | GOREGAON | KANDIVALI (E) | KANDIWALI (W) | BORIVALI | BHAYANDER | VASAI | POWAI | DADAR |SION | THANE | LOKPURAM (THANE) | DOMBIVLI | KALYAN | PANVEL | KAMOTHE | NERUL | SANPADA | KHARGHAR |

24. (C)Sol: U1 Un – Un – 1

2cos 2cos n 2 cos(n 1)

2 2cos cos n 2cos n 1

2 cos n 1 cos n 1 2 cos n 1

2cos n 1

= Un + 1Topic: Trigonometric Ratios_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_AGm sir.

25. (B)Sol: As a, b,c are roots of x3 – 15x2 + 47x – 82 = 0

a + b + c = 15 ab + bc + ac = 47 abc = 82

Now, cos A cos B cos Ca b c

2 2 2 2 2 2 2 2 2b c a a c b a b c

2abc 2abc 2abc

22 2 2 a b c 2 ab bc caa b c2abc 2abc

215 2 47 1312 82 164

Topic: Solution of Triangles_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_AGm sir.

26. (B)

Sol: We have 2 4 6 2x cos 3 ycos 16cos 9cos

23 4 2 6x 4cos 3cos ycos 9cos 16cos

6 4 2 4 2 6x 16cos 24cos 9cos ycos 9cos 16cos

2 4 69x 9 cos y 24x cos 16x 16 cos 0

Above equation is an identity cos 9x – 9 = 0 x = 1 y – 24x = 0 y = 24

Topic: Trigonometric Ratios_L-2_IIT-JEE_JR_Target-2016_MT-3_Main Level_AGm sir.



27. (C)

Sol: a b c b c a bc

2 2b c a bc

2 2 2b c 2bc a bc

2 2 2b c a 2 bc

2 2 2b c a 2

2bc 2

2cos A

2

A is angle of – 1 < cos A < 1

21 1

2

0 4 Topic: Solution of Triangle_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_AGm sir.

28. (A)Sol: Numerator of given expression (sin 8x + sin 6x) + 6(sin 6x + sin 4x) + 12 (sin 4x + sin 2x)

= 2 sin 7x cos x + 12 sin 5x cos x + 24 sin 3x cos x= 2 cos x (sin 7x + 6 sin 5x + 12 sin 3x)

sin8x 7sin 6x 18sin 4x 12sin 2x 2cos xsin 7x 6sin 5x 12sin 3x

Topic: Trigonometric Ratios_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_AGm sir.

29. (B)

Sol: 3 3 3 33 3 3

3 3 3 3 3 3

2R sin A sin B sin Ca b csin A sin B sin C sin A sin B sin C

= (2R)3 = 8 R = 1Greatest length of a side of a inscribed in a circle c an be equal to diameter of the circle, Hencemaximum value of a is 2.

Topic: Solution of Triangle_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_AGm sir.

30. (B)Sol: T _ _ _ _ _ _ _ _ _ T

No. of words 9! 90720

2!2!

Topic: Permutation and Combination_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_AGm sir.



PART-B: PHYSICS= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =31. (D)

Sol:PFt

Momentum transferred to the wall,1,P mv tn

1/mvF nmv

n

Topic: NLM _L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_Hp sir.

32. (D)Sol: BC is isochoric , ,B A B C D CV V V V V V Topic: Heat & Thermodynamics_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_Hp sir.

33. (A)

Sol: Here 21 2

12

P P P

or 21 22 /v P P

5 52 3.5 10 3 10 /1000

52 0.5 101000

or5

12 0.5 10 101000

ms

Topic: Fluid Mechanics_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_Hp sir.

34. (B)Sol: Moles in ‘A’ initially and finally will be same

( / 3) ''

PA L P AxRT RT

....(1)

Moles of ‘B’ remain same2 ( / 3) ' ( 2 )

'PA L P A L x

RT RT

....(2)

Divide (1) by (2) 1 / 42 2

x x LL x

Topic: Heat & Thermodynamics_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_Hp sir.



35. (A)

Sol:(mg + ma)

net accel. sin

sinm g a

g am


36. (B)Sol: Components of velocity before and after collision parallel to the plane are equal,So

sin 60º sin 30ºv u .....(1)Components of velocity normal to the plane are related to each other

cos60º (cos30º )v e u ....(2)

cot 60º cot 30ºe

cos60ºcot 30º

e

133

e

13

e .

Topic: System of particles_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_Hp sir.

37. (B)Sol: Mass of water = 1000 kg

1000 10 1.515

PE mgh

kJ

Topic: WEP _L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_Hp sir.

38. (A)Sol: /Y Stress Strain

Hence A is the correct answer.Topic: Elasticity_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_Hp sir.



39. (A)Sol: Let the total time of race be T seconds and the distance be 100S m .

The velocity vs time graph isArea of OAD

2 21 1 8 3 5 /5 5 2 3 2 3s s T Ta or T m s

Topic: Kinematics_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_Hp sir.

40. (B)

Sol:

sinF Mg MA

sinsinM M F MgF xg T xL L M

1 xT F NL

Topic: NLM_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_Hp sir.

41. (B)Sol: Let AB be a string of length L. Let F be the force pulling the rope to the right

Mass per unit length of rope ML

where M is the total mass.

P

xF

BALet P be a point at a distance x from B. If T is the tension in the rope at P then tension is towards rightwhile for the part PB it is towards left. If a is the acceleration of rope, then for part PB

F T mass of PB × aMxF T aL

Also for the rope, F Ma



F L xT

L


42. (C)

Sol:0.91 9

0.9 0.1x x x

9 80 720Q cal Topic: Heat & Thermodynamics_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_Hp sir.

43. (D)

Sol: Acceleration without friction sin2

gg

With friction, the acceleration is (sin cos ) (1 )2

gg

Since the body starts from rest, the distance is equal to 212

at .

Thus, 2 21 2(1 )

2 2g gS t t

Given 2 1t t

2

11

Topic: Friction_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_Hp sir.

44. (A)Sol: A BX X

2110.5 102

t at tan 45º 1a

2 20 21 0t t 20 400 842

t

21 sec.t Topic: Kinematics_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_Hp sir.

45. (C)Sol: Strain developed :

6 4(12 10 )(50) 6 10T Strain will be negative as the rod is in a compressed state.

6 4(12 10 )(50) 6 10T Topic: Elasticity_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_Hp sir.



46. (C)

Sol: 125 70500 40

X Y

For 501375.0º

YX X


47. (B)Sol: ConceptualTopic: WEP _L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_Hp sir.

48. (D)

Sol: 3buoyant wm mgF g

23 3applied

mg mgF mg

work done by applied force 2 3 100

3mg J

Topic: WEP_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_Hp sir.

49. (A)Sol: Let 5&w wV u V v

Time taken by swimmer to go from M to O and O to B=time taken by float to reach B from M

11 122

V u

v u u

1 2 12 2( )

v uv u u

2 12( )

v u v uv u u

(2 2) 2( )v u v u

2 2 2 2vu u v u 1 /u km h Topic: Kinematics_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_Hp sir.



50. (B)

Sol:

Tension at a distance x from right end is 1 xT FL

Extension in dx width of rod is 1T F xdy dx dxYA YA L

net extension 0 0

12

L L F x FLdy dxYA L YA

Topic: Elasticity_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_Hp sir.

51. (C)

Sol:2

sin mvT mgR

20 2 sin 30º

3 sin 30ºm u gl

mg mgl

mg sin 30º mg cos 30º

mg

T= 3 mg

0 3 / 2u g Topic: Circular motion_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_Hp sir.

52. (A)

Sol: Energy Density 2 31 1 ( ) 2880 /2 2

stress strain Y strai J m


53. (A)Sol: As long as the block of mass m remains stationary, the block of mass M released from rest comes

down by 2Mg

K (before coming it rest momentanly again).

Thus the maximum extension in spring is2Mgx

K .....(1)

for block of mass m to just move up the incline



sin coskx mg mg ....(2)

3 3 425 4 5

Mg mg mg or 35

M m

Topic: WEP _L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_Hp sir.

54. (B)

Sol:21 1

2 2M g h MV MS T

21 110 50 25 2 56.252 2

T

500 625 10º56.25 2

T C


55. (B)

Sol: Using 2 2 2 21 2 3 4

4rmsv v v vV

ans substituting the corresponding velocity we get rmsV is greatest in situation b.Topic: Heat & Thermodynamics_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_Hp sir.

56. (C)Sol: Let velocity of projection be V and velocity of the block when it returns back = 'V

then 'V V (since some K.E. is lost to friction)Hence average velocity during ascent > average velocity during descent

a dt t Topic: Kinematics_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_Hp sir.

57. (D)

Sol:

0 5 2 10v t t

01 11500 .3 10 .32 2

S V t t t

10sec.t total time = 3t = 30 sec.

Topic: Kinematics_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_Hp sir.



58. (C)

Sol: Strain 3( ) )(200) 2 10ll

Stress 11 3 8 210 2 10 2 10 /N m

Required force = stress × Area = 8 6 22 10 2 10 4 10 400 N

Mass to be attached 400 40kgg


59. (A)Sol: The tension vector T

is always normal to velocity vector. Hence .T v is always zero.

A is correct choice.Topic: Circular motion_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_Hp sir.

60. (B)Sol: Velocity of knife edge when it hits the floor

2v gh

Retardation, 2 2

2 2v gh ghas s s

Equation R Mg Ma

1gh hMg M Mgs s

Topic: Friction_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_Hp sir.



PART-C: CHEMISTRY= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =61. (A)Sol: Note Concentration of a substance in mol per liter is known as molarity.

Given, 13d 1.14g mL , mass %of HNO 69%

69% HNO3 means 100g of its solution contains 69 g HNO3 (nitric acid).Hence, mass of HNO3 (solute) = 69 gMolar mass of nitric acid

3HNO 1.0079 14.0067 3 16.00 163.0146g mol

Density, mdV

or 1

m 100g 100V mLd 1.41g mL 1.41

w 1000Molarity

m volumeof solution mL

69 1000 1.41 15.439M63.0146 100

Alternative Method

density mass percent 10Molaritymolar mass of solute

1.14 69 10 15.439 M

63.0146

Topic: Volumetric_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_SSch sir.

62. (C)

Sol: 1.53% of V 100 55.842.74

& %of O 44.16

Element Percentage Atomic ratio Simplest ratio

V 55.8455.84 1.1

52

1.1 11.1

O 44.1644.16 2.76

16

2.76 2.51.1

Ratio of V : O 2 : 5

Thus, empirical formula = V2O5

Topic: Mole Concept_L-2_IIT-JEE_JR_Target-2016_MT-3_Main Level_SSch sir.



63. (B)Sol: Suppose final temperature is t and Heat Q ms t

Heat lost by silver 100 0.0565 40 t

Heat gained by water 60 1 t 10

Heat lost by silver = Heat gained by water (As per first law)

5.65 40 t 60 t 10

t 12.58 C Topic: Thermodynamics_L-2_IIT-JEE_JR_Target-2016_MT-3_Main Level_SSch sir.

64. (D)Sol: From Gay-Lussac’s law,

1 2

1 2

p pT T

1 1.1t 273 t 283

t 283 1.1t 300.3 0.1t 17.3

t 173 C

t 273 173 100 K Topic: Gaseous State_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_SSch sir.

65. (C)Sol: The ionic radii of isoelectronic species decreases with increase in atomic number (as magnitude of

the nuclear charge increases with increase in atomic number). Therefore, that ionic radii increase inthe order

Isoelectronic ions 3 2 2 3Al Mg Na F O N

Atomic Number, Z 13 12 11 9 8 7Topic: Periodic Table_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_SSch sir.

66. (B)Sol: 1 mol14 C = 14 g = 6.022×1023 carbon atoms

Number of neutrons in 1 carbon atom= mass number – atomic number = 14 – 6 = 8 neutrons

6.023 × 1023 carbon atoms will contain 6.023×1023×8 neutrons 14 g carbon - 14 have 6.023×1023×8 neutrons and 7mg or 7×10–3 g carbon-14 will have

3 237 10 6.023 10 8 neutrons14

20 2124.092 10 neutrons 2.4092 10 neutrons Topic: Atomic Structure_L-2_IIT-JEE_JR_Target-2016_MT-3_Main Level_SSch sir.



67. (A)

Sol:2 2 2

2 2 2

N N SO

SO SO N

r T Mr T M

2NT1.625 641 323 28

2N

1.625 323 28T

64

373K

Topic: Gaseous State_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_SSch sir.

68. (B)Sol: CI has more negative electron gain enthalpy than fluorine. All other given statements are correct.Topic: Periodic Table _L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_SSch sir.

69. (A)

Sol: Mole fraction of 2nI 0.1

n N

...(i)

Mole fraction of benzene N 0.9

n N

...(ii)

where, n and N are moles of I2 and C6H6 in solution respectively.

molality,

2

2

1000 mole fraction of Im1 mole fraction of I M

1000 0.1 1.42 m

0.9 78


70. (C)

Sol: 12 8100pm 100 10 m 1 10 cm

(i) Volume of NA molecules, 3A

4V r N3

38 234 22 1 10 6.022 103 7

3 1cm mol

3 3 3 12.52cm 2.52 10 dm mol

3 3 1b 4V 4 2.52 10 dm mol

3 3 110.08 10 dm mol 2 3 11 10 dm mol




71. (D)Sol: KE = quantum energy – threshold energy

34 8 34 8

10 10

6.626 10 3 10 6.626 10 3 10KE3000 10 4000 10

19 196.626 10 4.969 10

191.6565 10 J Topic: Atomic Structure_L-2_IIT-JEE_JR_Target-2016_MT-3_Main Level_SSch sir.

72. (C)Sol: In all the given molecules, anion (i.e., F–) is common, thus lattice energy depends upon the size of

cations. The order of size is Na K Rb Cs

Lattice energy 1

sizeof ion

Thus, the order of lattice energy is NaF KF RbF CsF Topic: Chemical Bonding_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_SSch sir.

73. (D)

Sol: * *

2 2 2 2 2 2 22 x y zN 7 7 14 1s , 1s , 2s , 2s , 2p 2p , 2p

10 4BO 32

* *

2 2 2 2 2 2 12 x y zN 7 7 1 13 1s 1s , 2s , 2s , 2p 2p , 2p (paramagetic)

9 4BO 2.52

* *

2 2 2 2 22 zO 8 8 16 1s , 1s , 2s , 2s , 2p ,

* *

2 2 1 1z y x y2p 2p , 2p 2p

10 6BO 22

2O 8 8 1 15 10 5BO 2.5

2

Thus, 2 2 2 2N N and O O (stability order)

2N is paramagneticTopic: Chemical bonding_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_SSch sir.



74. (B)

Sol: Configuration in second excited state is written as : 2 3 2Ne 3s 3p 3d

3p

3d

nlm

3 3 31 1 11 0 1

3 32 22 1

Total n l m 25

Topic: Atomic Structure_L-2_IIT-JEE_JR_Target-2016_MT-3_Main Level_SSch sir.

75. (B)Sol: The following steps are involved to balance this equation.

(i) Split the equation into oxidation and reduction half - reactions.2

4MnO Mn ; 2

2 4 2C O CO (ii) Balance the atoms other than O and H.

24MnO Mn ;

22 4 2C O 2CO

(iii) Add H2O to the side deficient in O and H+ to the side deficient in H. 2

4 2MnO 8H Mn 4H O;

22 4 2C O 2CO

(iv) Balance the charge by adding electrons.2

4 2MnO 8H 5e Mn 4H O

22 4 2C O 2CO 2e

(v) Equalise the number of electrons and then add both the equations.2

4 2MnO 8H 5e Mn 4H O 2

22 4 2C O 2CO 2e 5

2 24 2 4 2 22MnO 5C O 16H 2Mn 10CO 8H O

Thus, the coefficients of 24 2 4MnO ,C O and H are respectively 2, 5 and 16.

Topic: Redox_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_SSch sir.

76. (B)Sol: E.C. of O2 (16 electrons)

* * * *2 2 2 2 2 2 2 1 1

z x y x y1s , 1s , 2s , 2s , 2p , 2p 2p , 2p 2p

Bond order b a1 1N N 10 6 22 2

E.C. of 2O (15 electrons)



* * * *2 2 2 2 2 2 2 1

z x y x y1s , 1s , 2s , 2s , 2p , 2p 2p , 2p , 2p

Bond order = b a1 1N N 10 5 2.52 2

2E.C.of O 17 electrons

* *2 2 2 2 2 2 2 2 1

z x y x y1s , 1s , 2s , 2s , 2p , 2p 2p , 2p 2p

Bond order b a1 1N N 10 7 1.52 2

Thus, the order of bond order is 2 2 2O O O Topic: Chemical Bonding_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_SSch sir.

77. (B)

Sol:x 2

4 3 215M 2MnO MO Mn O2

+7 +5

Change in oxidation state = 2

Change in oxidation state = 5

Oxidation

Reduction x 2 5

x 5 2 3 Topic: Redox_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_SSch sir.

78. (C)Suppose volume of CO in the mixture = x ml

Then volume of N2 in the mixture = 10 x ml

Reaction (i) 2 212g g gCO O CO

Reaction (ii) 2 2 2g g gN O NO

O2 required for reaction (i) = 2x

O2 required for reaction (i) = 10 x ml

Now 10 72x x x = 6ml

Hence, volume of N2 in the mixture = 10 6 4ml Topic: Mole concept-1; Sub Topic: Eudiometry: L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_SSch sir.



79. (B)Sol: (i) Since, in it the last electron enters in s-orbital, it is a s-block element and s-block elements are

good reducing agents.(ii) Since, the last electron enters in f-block, it is a f-block element (element of 4f series).(iii) Electronic configuration reveals that it contains 7 electrons in its outer shell, so it is a halogen and

and halogens have high magnitude of ge H .

(iv) The last electron enters in d-orbital, so it is a d-block element and variable oxidation states is animportant feature of d-block elements.

Topic: Periodic Table_L-2_IIT-JEE_JR_Target-2016_MT-3_Main Level_SSch sir.

80. (C)Sol: A single bond 1 bond

Double bond = 1 and1 bond

Triple bond 1 and 2 bonds

Thus, the given molecule contains 19 and 5 bonds Topic: Chemical bonding_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_SSch sir.

81. (C)

Sol: 2 2 31 3X Y XY2 2

reaction products reactantsS S S

1 1reaction

3 1S 50 40 60 40J mol K2 2

G H T S At equilibrium as G 0 H T S

3H 30 10T 750K

S 40

Topic: Thermodynamics_L-2_IIT-JEE_JR_Target-2016_MT-3_Main Level_SSch sir.

82. (B)

Sol: tNo :of moles Molarity Volume Lmol.wt

0.375 82 500wt 15.38g

1000




83. (B)Sol: Moalr mass of CO2 gas = 44g/molTopic: Mole concept_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_SSch sir.

84. (A)

Sol:p

av

2RTV MV 28RT

M


85. (B)

Sol: 3m 10g 10 10 kg

Uncertainty in speed 1 14 90v 4% of 90 ms 3.6ms100

From Heisenberg uncertainty principle,h hx v or x

4 m 4 m v

Uncetainty in position,

34 2 1

3 1

6.626 10 kgm sx4 3.14 10 10 kg 3.6 ms

331.46 10 m Topic: Atomic Structure_L-2_IIT-JEE_JR_Target-2016_MT-3_Main Level_SSch sir.

86. (B)Sol: Melting point depends upon lattice energy. NaCl and KCl have unit charge on their ions while MgO

and BaO have two units of charge, therefore, lattice energies of MgO and BaO are expected to belarger than those of NaCl and KCl.Since, Mg2+ is smaller than Ba2+, therefore, MgO has the highest lattice energy and hence, has thehighest melting point.

Topic: Chemical Bonding_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_SSch sir.

87. (B)Sol: For H2O2 solution,

Volume strength = 5.6 × Normality

Normality of 2 220H O 3.575.6




88. (C)

Sol: 2 21 1I s Cl g ICl g2 2

s g diss 2 diss 2 ICl1 1 1H H H Cl H I H2 2 2

1 1 162.76 242.3 151.0 211.32 2 2

228.03 211.3 16.73 kJmol–1.


89. (B)

Sol: gn for the equilibirum reaction 11 1·5 0 52

1ng2

p c cK K RT K RT

c

pKKRT

Topic: Chemical Equilibrium_L-1_IIT-JEE_JR_Target-2016_MT-3_Main Level_SSch sir.

90. (A)Sol: 1

vap bH 40.63kJ mol , T 373K 1

vapvap

b

H 40.63 1000 J molST 373K

1 1109 JK mol

Entropy change for evaporation of 36 g of water 1109 36 218JK18


RIIT Major Test -3 Mains Solutions (18!01!2015) (1)

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